Question

A tank initially contains 1000 liters of salt solution with 70 grams of salt per liter. A solution containing 120 grams of salt per liter enters the tank at the rate of 9 liters per minute, and the well mixed solution is pumped out at 10 liters per minute. Find the concentration of salt as a function of time.

Answer 1

In (1080 - 10C) = 10 In (1079.3 - 1.0793t)

Explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time

The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)

The rate of change of the volume of solution = dV/dt

Rate of flow into the tank = Fᵢ = 9 L/min

Rate of flow out of the tank = F = 10 L/min

dV/dt = Fᵢ - F

dV/dt = (Fᵢ - F)

dV = (Fᵢ - F) dt

∫ dv = ∫ (Fᵢ - F) dt

Integrating the left hand side from 1000 litres (initial volume) to V and the right hand side from 0 to t

V - 1000 = (Fᵢ - F)t

V = 1000 + (Fᵢ - F)t

V = 1000 + (9 - 10) t

V = 1000 - t

Component balance for the concentration.

Let the initial concentration of salt in the tank be C₀ = 70/1000 = 0.07 g/L

Let the amount of salt coming into the tank be 120 g/L × 9 L/min = 1080 g/min

Concentration coming into the tank = Cᵢ = 1080/V (g/L.min)

Concentration of salt in the tank, at any time = C g/L

Rate at which that Concentration of salt is leaving the tank = (C × 10 L/min)/V = (10C/V) g/L.min

Rate of Change in the concentration of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)

Rate of Change in the concentration of salt in the tank = dC/dt

Rate of flow of salt into the tank = (1080/V) g/L.min

Rate of flow of salt out of the tank = (10C/V) g/L.min

(dC/dt) = (1080/V) - (10C/V) = (1080 - 10C)/V

Recall, V = 1000 - t

(dC/dt) = (1080 - 10C)/(1000 - t)

dC/(1080 - 10C) = dt/(1000 - t)

∫ dC/(1080 - 10C) = ∫ dt/(1000 - t)

Integrating the left hand side from C₀ (initial concentration of salt) to C and the right hand side from 0 to t

- 0.1 In [(1080 - 10C)/(1080 - 10C₀)] = - In [(1000-t)] + K (where k = constant of integration)

At t= 0, C = C₀

- 0.1 In 1 = (- In 1000) + K

0 = (- In 1000) + K

K = In 1000

- 0.1 In [(1080 - 10C)/(1080 - 10C₀)] = - In [(1000-t)] + In 1000

0.1 In [(1080 - 10C)/(1080 - 10C₀)] = In [(1000-t)] - In 1000

In [(1080 - 10C)/(1080 - 10C₀)] = 10 In [(1000-t)/1000]

In (1080 - 10C) - In (1080 - 10C₀) = 10 In [(1000-t)/1000)

In (1080 - 10C) = 10 In [(1000-t)/1000)] + In (1080 - 10C₀)

C₀ = 0.07 g/L

In (1080 - 10C) = 10 In [(1000-t)/1000)] + In (1080 - 10(0.07))

In (1080 - 10C) = 10 In [(1000-t)/1000)] + In (1080 - 0.7)

In (1080 - 10C) = 10 In [(1000-t)/1000)] + In (1079.3)

In (1080 - 10C) = 10 In [1.0793(1000-t)]

In (1080 - 10C) = 10 In (1079.3 - 1.0793t)

Answer 2

10 - A(t)/(1000 + t)

Explanation:

salt enters the tank at the rate of 9 liters per minute, and is pumped out at 10 liters per minute.

`\fracdv}{dt} = 10-9 = 1 liters/mins`

V(0) = 1000 liters

V(t) = 1000 + t.

Let A(t) be the amount of salt in gram at time t

The concentration of salt entering the tank is 120 grams/liters.

`(A(t))/(v(t))` grams/liters is the concentration of the salt that is being pumped out of the tank at time t

`(da)/(dt)` = 1 * 10 - A(t)/(1000 + t)

= 10 - A(t)/(1000 + t)